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Q.

If  tanθ+sinθ=m  and  tanθsinθ=n,  then 

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a

m2+n2=4mn

b

m2n2=4mn

c

m2n2=4mn

d

m2n2=m2+n2

answer is D.

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Detailed Solution

tanθ+sinθ=mtanθsinθ=n
mn=tan2θsin2θ =tan2θ(1-cos2θ) =tan2θ sin2θ
m2-n2=(tanθ+sinθ)2-(tanθ-sinθ)2 =4tanθsinθ m2-n2=4mn

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